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ELECTRONS IN ATOMS

This article covers the fundamental concepts of electrons in atoms, including wave properties, electromagnetic radiation, atomic spectra, Planck's quantum theory, and the Bohr atom model.

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ELECTRONS IN ATOMS

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  1. ELECTRONS IN ATOMS Dr Mehmet Gökhan ÇAĞLAYAN 09/2019

  2. Basic concepts • A wave is a disturbance that transmitsenergy through space or a material medium. • The distancebetween identical points on successive waves is called the wavelength λ(lambda). • The frequency v (nu) of the wave is the number of waves that pass througha particular point in one second. • The amplitude is the vertical distance from themidline of a wave to the peak or trough. • Electromagneticradiationis the emission and transmission of energy in the form of electromagnetic waves.

  3. Speed of thewave speed = wavelength * frequency u = λ * v • Wavelength is usually expressed in units of meters, centimeters, or nanometers, • frequency is measured in hertz (Hz); 1 Hz = 1 cycle/s = 1 s-1

  4. The wavelength of electromagnetic radiation is shorter forhigh frequencies and longer for low frequencies. • The visible region, extends from violet, 400nm, to red, 780 nm. • Gamma rays have the shortest wavelength and highest frequency; radio waves have thelongest wavelength and the lowest frequency Theelectromagneticspectrum

  5. Question • Most of the light from a sodium vapor lamp has a wavelength of 589 nm. What is the frequency of this radiation? Change the units of from nanometers to meters. v= c/ λ = 2.998* 108 m s-1 / 5.89 * 10-7 m = 5.09 * 1014 Hz

  6. Question • From the following colors of visible light, which is most energetic? (a) green, (b) red, (c) yellow. the higher the frequency, the more energetic the radiation. GREEN YELLOW RED

  7. Characteristic of ElectromagneticWaves

  8. TheVisibleSpectrum The spectrum of white light • (a) Dispersion of light through a prism. Red light is refracted the least and violet lightthe most when white light is passed through a glass prism. The other colors of thevisible spectrum are found between red and violet. • (b) Rainbow near a waterfall.Here, water droplets are the dispersion medium.

  9. AtomicSpectra • The visible spectrum is a continuous spectrum because thelight being diffracted consists of many wavelength components. If the source ofa spectrum produces light having only a relatively small number of wavelengthcomponents, then a discontinuous spectrum is observed. • Ifthelightsource is an electric discharge passing through a gas, only certain colors are seenin the spectrum. • Or, if the light source is a gas flame into whichan ionic compound has been introduced, the flame may acquire a distinctivecolor indicative of the metal ion present. These discontinuous spectra are called atomic, or line, spectra

  10. Planck’s Quantum Theory • Max Planck (1858 1947) made a revolutionary proposal: Energy, like matter, is discontinuous. • Classical physics places nolimitations on the amount of energy a system may possess, whereas quantumtheory limits this energy to a discrete set of specific values. • Planck gave the namequantumto the smallest quantity of energy thatcan be emitted (or absorbed) in the form of electromagnetic radiation. E = hv h = 6.62607 * 10-34 J s (Planck’sconstant) v= c/ λ E = h c /λ • According to quantum theory, energy is always emitted in integral multiples ofhv; for example, hv, 2 hv, 3 hv

  11. Question • Calculate the energy (in joules) of (a) a photon with a wavelength of 3.75 * 104 nm(infrared region) and (b) a photon with a wavelength of 3.75 * 10-2 nm (X ray region). a) E = h c /λ =(6.63 * 10-34 J s)(3.00 * 108 m/s) /(3.75 * 104 nm) / (1*10-9 m/1 nm) = 5.30* 10 -21 J b) 5.30* 10 -15 J

  12. Question • For radiation of wavelength 242.4 nm, the longest wavelength that will bring about the photodissociation ofwhat is the energy of (a) one photon, and (b) a mole of photons of this light? a) v = c/ λ = 2.998 * 108 m s-1/ 242.4 * 10-9 m = 1.237 * 1015s-1 E = hv= 8.196 * 10-19 J/photon b) E = 8.196 * 10-19 J/photon * 6.022*1023 photons/mol

  13. Homework 1) The protective action of ozone in the atmosphere comes through ozones absorption ofUV radiation in the 230 to 290 nm wavelength range. What is the energy, in kilojoules per mole, associatedwith radiation in this wavelength range? 2) Chlorophyll absorbs light at energies of 3.056 * 10-19 J/photonand 4.414 * 10-19 J/photon. To what color and frequency do these absorptions correspond?

  14. Bohr Atom • Acc. to the laws of classical physics, an e moving in an orbit of ahydrogen atom would experience an acceleration toward the nucleus by radiating awayenergy in the form of electromagnetic waves. Thus, such an ewould quicklyspiral into the nucleus and annihilate itself with the p. • why this doesnot happen: Bohr postulated that the e is allowed to occupy only certain orbitsof specific energies. Ane in any of the allowed orbits will not radiate energy, sowill notspiral into the nucleus. • Bohrshowed that the energies that anein hydrogen atom can occupy are given by • where RH, the Rydberg constant for the hydrogen atom = 2.18*10-18 J. • The number n is an integer calledthe principal quantum number; it has the values n 5 1, 2, 3, . . . .

  15. Bohr model of H atom The nucleus is at thecenter, and the ein one of the discrete orbits. Excitation of the atom raises the electron tohigher-numbered orbits, as shown with black arrows. Light isemitted when the electron falls to a lower-numbered orbit. Twotransitions that produce lines in the Balmer series of the hydrogenspectrum are shown in the approximate colors of the spectral lines.

  16. Transitions in the hydrogen spectrum • The energy levels in the hydrogen atom. • Each energy levelcorresponds to the energy associated with an allowed energy state for an orbit.

  17. Energy-leveldiagramforthehydrogen atom

  18. Energydifference • Ground state is thethe ground lowest energy state of an atom • The stability ofthe electron diminishes for n = 2, 3, . . . Each of these levels is called an excitedstate, or excited level, which is higher in energy than the ground state. • A hydrogenelectron for which n is greater than 1 is said to be in an excited state. • During emission, the electron drops to a lower energy state : nfinal (nf). The difference between the energies of the initial (ni)and final states is: • ∆ E =Ef- Ei

  19. Question • Determine the wavelength of the line in the Balmer series of hydrogen corresponding to the transition from n=5 to n=2? ∆ E = 2.179 * 10-18 J(1/52-1/22)= -4.576*10-19 J Ephoton= ∆ E= hv v= Ephoton/h = 4.576*10-19 Jphoton-1 /6.626*10-34 Jsphoton-1 =6.906 * 1014 s-1 λ =c/v =434.1 nm

  20. Homework • What is the wavelength (in nanometers) of a photon emitted duringa transition from ni= 6 to nf= 4 state in the H atom?

  21. Waveproperties • De Broglie’s contribution: waves can behave like particlesand particles can exhibit wavelike properties. λ =h/mu λ, m, and u are the wavelengths associated with a moving particle, its mass,and its velocity, respectively. Equation implies that a particle in motion canbe treated as a wave, and a wave can exhibit the properties of a particle.

  22. Question Calculate the wavelength of the “particle” in the following two cases: (a) Calculatethe wavelength associated with a 5.50 x 10-2kg tennis ball traveling at 45 m/s.(b) Calculate the wavelength associated with an electron (9.1094 x 10-31 kg) moving at 45 m/s. • λ =h/mu = 2.7*10-34 m • 1.6 * 10-5 m

  23. Ionization Energy of Hydrogen • The Bohr model of the atom helps to clarify the mechanism of formation ofcations. • The model also works for hydrogen-like species, such as the ions He+2, Li+2which have only one electron.

  24. Question • Determine the kinetic energy of the electron ionized from a Li+2 ion in its ground state, using a photon offrequency 5.000*1016 s-1. Ephoton = IE +Keelectron

  25. Quantum NumbersandElectronOrbitals • 3 quantum numbersare derived from themathematical solution of the Schrödinger equation for the hydrogen atom. • Thesenumbersrequiredto describe the distribution of electrons in hydrogen and other atoms. • n : principalquantumnumber; determines the energy of an orbital; 1,2,3.. • l :theangular momentum quantumnumber; determines the “shape” of the orbitals, 0, 1,2..n-1 • ml :themagneticquantumnumber; describes the orientation of the orbital in space-l, (-l +1)….0…(l-1),+l

  26. Quantum numbers • A fourthquantumnumber—the spin quantum number—describes the behavior of a specific electron and completesthedescription of e. • ms= +1/2 or -1/2

  27. Quantum Numbers and Atomic Orbitals

  28. the hydrogen 1s, 2s, and 3sorbitals surfacediagramsof the three 2p orbitals Boundary surface diagrams of the five 3d orbitals

  29. Question • What is the total number of orbitals associated with the principal quantum number n = 3? • For n=3, the possible values of l= 0, 1,2. • one 3sorbital (n=3, , l= 0, and ml= 0); • three 3p orbitals (n=3, , l=1, and ml=-1,0,1); • five 3d orbitals (n=3, , l=2, and ml=-2,-1,0,1,2). • The total number of orbitals is 1 + 3 + 5 = 9.

  30. Question • From the following sets of quantum numbers (n,l,ml,ms)identify the set that is correct, and state the orbitaldesignation for those quantum numbers:

  31. Question • Write the four quantum numbers for an electron in a 3p orbital n, l, ml, ms

  32. Rules for Assigning Electrons to Atomic Orbitals • Each shell or principal level of quantum number n contains n subshells. Ex: n =2, then there are two subshells of angularmomentum quantum numbers (l) 0 and 1. 2. Each subshell of quantum number l contains (2l +1) orbitals. Ex: l=1 then there are three p orbitals. 3. No more than two electrons can be placed in each orbital. So, the maxnumberof electrons is simply twice the number of orbitals that areemployed. 4. the maximum number of electrons that an atom canhave in a principal level n can be foundbythe formula 2n2.

  33. Question What is the maximum number of electrons that can be present in the principal level forwhichn=3? n =3, l= 0, 1, and 2 2*9 orbitals= 18 e-s. Check: 2n2 =18

  34. Energies of Orbitals Orbital energy levels in thehydrogen atom. Orbital energy levels in amany-electron atom.

  35. ElectronConfiguration Electronconfig. Forgroundstate e of hydrogen atom Orbitaldiagram

  36. Multielectronatoms • The shielding effect refers to the effectof inner-shell electrons in shielding orscreeningouter-shellelectronsfromthe full effects of the nuclear charge.In effect the inner electrons partiallyreducethenuclearcharge. • Radial probability plots for the1s, 2s, and 2p orbitals. The 1selectronseffectivelyshieldboththe 2s and 2p electrons from thenucleus. The 2s orbital is morepenetrating than the 2p orbital.

  37. ThePauliExclusionPrinciple • For many-electron atoms, theprinciplestateselectrons in an atom should have the same n,l, ml andms values (that is, these two electronsare in the same atomic orbital), so they must have different values of ms. • Only two electrons may occupy the same atomic orbital, and these electronsmusthaveoppositespins.

  38. Hund’sRule Forcarbon (Z = 6) is 1s22s22p2whichconfig. is true? exclusionprinciple Acc. toPaulie’sprincipleall of themaretrue. However; the answer is provided by Hund’srule; statingthat the most stablearrangement of electrons in subshells is the one with the greatest number of parallelspins. C is thetrueanswer.

  39. Summary of rulesforelectronconfiguration • Electrons occupy orbitals in a way that minimizes the energy of the atom. 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p 2) No two electrons in an atom can have all four quantum numbers alike (thePauliexclusionprinciple). 3) When orbitals of identical energy (degenerate orbitals) are available,electrons initially occupy these orbitals singly (Hund’srule).

  40. Keep in mind

  41. DiamagnetismandParamagnetism • Paramagnetic substances are those that contain net unpairedspins and are attracted by a magnet. On the other hand, if the electron spins arepaired, or antiparallel to each other, the magnetic effects cancel out. • Diamagnetic substances do not contain net unpaired spins and are slightlyrepelled by a magnet. • Li atom contains one unpaired e and the Li metal is thereforeparamagnetic.

  42. TheAufbauProcess • Aufbauprocess (Germanword: building up) isto assign electronconfigurations to the elements of periodictable in order of increasing atomic number. • Electrons added to theshell of highest quantum number in the aufbau process arecalledvalence electrons.

  43. Electronconfig of someatoms

  44. Electron configurations and the periodic table

  45. Question Write (a) the electron configuration of mercury, and (b) an orbital diagram for the electron configuration of tin. • [Xe]4f145d106s2

  46. Question • Write the ground-state electron configurations for (a) sulfur (S) and (b) palladium (Pd),which is diamagnetic. • Sulfur has 16 electrons. The noble gas core in this case is [Ne]. 1s22s22p63s23p4or [Ne]3s23p4. b) Palladiumhas 46 electrons. The noble-gas core in this case is [Kr]. [Kr] represents 1s22s22p63s23p64s23d104p6 Palladium: 1s22s22p63s23p64s23d104p64d10 Answer: [Kr]4d10

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